If a local morphism of rings induces an isomorphism on associated graded rings, must it have been an isomorphism I have local morphism of local rings $f:(R,\mathfrak{m}) \to (S,\mathfrak{n})$, (i.e. $f(\mathfrak{m}^r) \subset \mathfrak{n}^s$ for all $s$) such that $\mathrm{gr}(f):\mathrm{gr}_{\mathfrak{m}}(R) = R/\mathfrak{m} \oplus \mathfrak{m}/\mathfrak{m}^2 \oplus \cdots \to \mathrm{gr}_{\mathfrak{n}}(S)$ is an isomorphism of graded rings. Can I conclude $f$ was an isomorphism to begin with? I feel like this should be clear one way or the other but I don't see it. Rings are not Noetherian but can be assumed complete if that matters/makes it true. If not true, are there any general conditions on the rings that would make this hold?
 A: If you drop the completeness requirement, you have the inclusion: $$\mathbb{R}\hookrightarrow\frac{\mathbb{R}[t_\lambda|\lambda\in \mathbb{R}^{>0}]}{\langle t_\lambda t_\mu- t_{\lambda+\mu}|\lambda,\mu\in \mathbb{R}^{>0},t_\lambda |\lambda\geq1 \rangle}$$
Note the RHS is a quotient of a polynomial ring (not formal power series).  The maximal ideal $m$ is generated by the $t_\lambda$.  To see see it is unique, note that all its elements are nilpotent, so any element outside of $m$ has the form $\lambda+\alpha$ with $\lambda\in \mathbb{R}, \lambda\neq0$ and $\alpha$ nilpotent, so $\lambda+\alpha$ is invertible.
Finally note that $m^2=m$ as $t_\lambda={t_{\frac\lambda2}}^2$.  Thus applying ${\rm gr}_m$ to the RHS just returns $\mathbb{R}$.
For any Cauchy sequence in the RHS, there exists $\lambda\in \mathbb{R}$ such that eventually all elements $x_j$ of the sequence have the form $\lambda+\alpha_j$, with $\alpha_j\in m$.  Then we have $\alpha_j\in m^i$ for all $i$ so the sequence converges to $\lambda$.  However Cauchy sequences converge to more than one limit.

Let $(R,\mathfrak{m})$ be a local ring.  Define a sequence in $R$ to be Cauchy if for all $n\in \mathbb{N}$, its residues modulo $\mathfrak{m}^n$ are eventually constant.  Say that $R$ is $complete$ if every Cauchy sequence has a unique limit.
Lemma: If $R$ is complete and $f\colon (R,\mathfrak{m}) \to (S,\mathfrak{n})$ is a local morphism, and
gr$(f)$ is injective, then $f$ is injective.
Proof: gr$(f)$ being injective means precisely that for all $i\in \mathbb{N}$, if $f(x)\in \mathfrak{n}^i$ then $x\in \mathfrak{m}^i$. In particular if $f(x)=0$ then $x\in\mathfrak{m}^i$ for all $i$ and the constant sequence $x,x,x,\cdots$ converges to $0$.  However it also converges to $x$, so as $R$ is complete, we have $x=0$. $[]$
Lemma: If $R,S$ are both complete and $f\colon (R,\mathfrak{m}) \to (S,\mathfrak{n})$ is a local morphism, and gr$(f)$ is surjective, then $f$ is surjective.
Proof: If gr$(f)$ is surjective, then for any $y\in S$ we can find $x_0\in R$ with $y-f(x_0)\in \mathfrak{n}$. Then we may find $x_1\in \mathfrak{m}$ so that $y-f(x_0)-f(x_1)\in \mathfrak{n}^2$.  Thus $y-f(x_0+x_1)\in \mathfrak{n}^2$. Repeating we may find $x_2\in \mathfrak{m}^2$ so that $y-f(x_0+x_1+x_2)\in\mathfrak{n}^3$, and so on.
The sequence $x_0,x_0+x_1,x_0+x_1+x_2,\cdots$ is Cauchy so converges to $x\in R$.  Thus for all $i\in\mathbb{N}$ we have $y-f(x)\in \mathfrak{n}^i$.  In particular, the constant sequence $f(x),f(x),f(x),\cdots$ converges to $y$.  However it also converges to $f(x)$ so as $S$ is complete we have that $y=f(x)$.[]
Combining these results, we have that if $R,S$ are complete, then gr$(f)$ an isomorphism implies that $f$ is an isomorphism.
